I just wonder how you would find $\mathrm P(y_1+y_2<1)$ given $0 \leq y_2 \leq y_1 < \infty$
Their joint density function is
$$f(y_1,y_2)=\begin{cases} e^{-y_1} ,& 0\leq y_2\leq y_1 < \infty\\[1ex] 0 ,& \textrm{otherwise}\end{cases}$$
I solved it by integrating $y_1$ over $y_2$ to $1-y_2$ then integrating $y_2$ over $0$ to $1$. I got $0$, which I found strange (and wrong?).
Looking forward to hearing from you! Thank you so much.
Good attempt, but you need
$$\int_0^{1/2}\int_{y_2}^{1-y_2} e^{-y_2}\operatorname d y_1\operatorname d y_2 ~=~ 1-2e^{-1/2}$$
Because since $y_2<y_1$ when $y_2>1/2$ then $y_2+y_1>1$